如何在日期时间聚合 percentile_disc() 函数

Question

我有如下表格：

recorddate score 
2021-05-01   0
2021-05-01   1 
2021-05-01   2
2021-05-02   3
2021-05-02   4
2021-05-03   5
2021-05-07   6

并且想要每周 score 获得第 60 个百分位数。我试过了：

select distinct 
       recorddate
     , PERCENTILE_disc(0.60) WITHIN GROUP (ORDER BY score)
                             OVER (PARTITION BY recorddate) AS top60
from tbl;

它返回了这样的东西：

recorddate top60
2021-05-01  1
2021-05-02  4
2021-05-03  5
2021-05-07  6

但我想要的结果就像是每周汇总（7 天）。例如，对于 2021-05-07 结束的一周：

recorddate                    top60
2021-05-01 ~ 2021-05-07        2

有解决办法吗？

Answer 1

我想你想要这个：

SELECT date_trunc('week', recorddate) AS week
     , percentile_disc(0.60) WITHIN GROUP(ORDER BY score) AS top60
FROM   tbl
GROUP  BY 1;

这是每周（存在实际数据）第 60 个百分位数的离散值 - 其中同一组（一周内）中 60% 的行相同或更小。准确的说，用the manual的话来说：

the first value within the ordered set of aggregated argument values whose position in the ordering equals or exceeds the specified fraction.

在上面添加您的格式：

SELECT to_char(week_start, 'YYYY-MM-DD" ~ "')
    || to_char(week_start + interval '6 days', 'YYYY-MM-DD') AS week
     , top60
FROM  (
   SELECT date_trunc('week', recorddate) AS week_start
        , percentile_disc(0.60) WITHIN GROUP(ORDER BY score) AS top60
   FROM   tbl
   GROUP  BY 1
   ) sub;

我宁愿把它叫做“percentile_60”。

如何在日期时间聚合 percentile_disc() 函数

How to aggregate percentile_disc() function over date time

sql

postgresql

datetime

aggregate-functions

percentile